不定积分∫(1/x^2)√(1-x)/(1+x)dx如何解?

∫√[(1-x)/(1+x)] dx/x^2x=cosu dx=sinu √[(1-x)/(1+x)]=(1-cosu)/sinu原式=∫(1-cosu)du/(cosu)^2=∫du/(cosu)^2-∫cosudu/(1-sinu)(1+sinu)=tanu-∫dsinu/[(1-sinu)(1+sinu)]=tanu-ln[|1+sinu|/|cosu|] +C=√[(1...